Complex Scalar Field in Terms of Two Independent Real Fields

  1. Oct 22, 2005 #1
    I am working with a complex scalar field written in terms of two independent real scalar fields and trying to derive the commutator relations.

    So,

    [tex] \phi = \frac{1}{\sqrt{2}} \left(\phi_1 + i \phi_2) [/tex]

    where [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are real.

    When deriving,

    [tex] [\phi(\vec{x},t),\dot{\phi}(\vec{x}',t)] = 0 [/tex]

    I get terms like the following:

    [tex][\phi_1(\vec{x},t),\dot{\phi}_2(\vec{x}',t)][/tex]

    which I need to vanish. It makes sense to me that they should vanish, but how do I show this?
     
  2. jcsd
  3. Oct 23, 2005 #2
    Hmm...I think that we just take that as the quantization condition. That is,

    [tex]
    [\phi_r(\vec{x},t),\pi_s(\vec{x}{\,}',t}] = i \delta^3(\vec{x}-\vec{x}{\,}')\delta_{rs}
    [/tex]

    Is this correct?
     
  4. Oct 23, 2005 #3

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Since [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are independent, they'll only be canonically conjugate with their own momenta (the [itex]\delta_{rs}[/itex] on the left). Your equation just states that in combination with the usual commutation relation of the real scalar field.
     
  5. Oct 25, 2005 #4
    [tex]\phi_1[/tex] and [tex]\phi_2[/tex]
    are independent fields, so
    [tex][\phi_1, \dot{\phi}_2][/tex]=0
     
  6. Oct 26, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    What is the Poisson bracket between the classical fields ? If you know that, you can canonically quantize using Dirac's rule.

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Complex Scalar Field in Terms of Two Independent Real Fields
  1. Complex scalar field (Replies: 2)

Loading...